Math 251 Spring 2004 Midterm Exam I1. Both f(t, y) =1y − 3t2and∂f∂y=−1(y − 3t2)2have to be continuous. The only conditionneeded is that y0− 3t206= 02. a) The equilibrium solutions are y = −3, 0, 1.b) y = −3 is unstable, y = 0 is (asymptotically) stable, and y = 1 is unstable.c) limt→∞y(t) = ∞d) limt→∞y(t) = 03. a) M(x, y) = y2exy2+ 4x3, N(x, y) = 2xyexy2+ 2∂M∂y= 2xy3exy2+ 2yexy2=∂N∂xSo the equation is an exact equation.b) exy2+ x4+ 2y = 54. a) The roots of the characteristic equation r2− 4r − 5 = 0 are r = −1, 5. Therefore,the functions y1= e−tand y2= e5tform a pair of fundamental solutions for this second orderlinear differential equation.b) General solution is y(t) = C1y1+ C2y2= C1e−t+ C2e5t.c) y(t) =76e−t+56e5t5. a) (Solved as a separable equation.) ey= sin t + t + e3(implicit solution), ory = ln(sin t + t + e3) (explicit solution).b) (Solved as a first order linear equation.) y(t) =2 ln |t|t+2t.6. a) The roots of the characteristic equation are r = −1 ±√2ı, the general solution istherefore y(t) = C1e−tcos√2t + C2e−tsin√2t.b) The root of the characteristic equation is r = −3 (repeated), the general solution istherefore y(t) = C1e−3t+ C2te−3t.7. a) non-linear; b) linear; c) non-linear; d) non-linear; e) linear.8. a) yes; b) no; c) no.9. First identify that p(t) =−4t; and t0= 1 so the solution interval is (0, ∞).By Abel’s Theorem, W (y1, y2) = Ce−Rp(t) dt= CeR4tdt= Ce4 ln t= Ct4.Calculate the Wronskian determinanty1y2y01y02at t = 1, W (y1(1), y2(1)) =1 20 3= 3, andequating it with the result obtained by Abel’s Theorem: 3 = C(14), hence C = 3. Therefore,W (y1, y2) = 3t4.110. Q(t) = 120γ − 120γe−160t. The limiting amount of salt is limt→∞Q(t) = 120γ (grams).(The limiting concentration would be γ grams/liter, but that is not what the question
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